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gavmur [86]
3 years ago
13

What is the maximum volume of 0.25 M sodium hypochlorite solution (NaOCl, laundry bleach) that can be prepared by dilution of 1.

00 L of 0.80 M NaOCl?
Chemistry
2 answers:
ehidna [41]3 years ago
8 0

Answer:

0.313L

Explanation:

0.25Mx1.00L=0.80Mx V2

so, 0.25x1.00/0.80=0.313L

Iteru [2.4K]3 years ago
7 0

Answer: 3.2L

Explanation:

C1 = 0.80M

V1 = 1L

C2 = 0.25M

V2 =?

C1V1 = C2V2

0.8 x 1 = 0.25 x V2

V2 = (0.8 x 1) /0.25 = 3.2L

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lubasha [3.4K]

For all three questions, we will use the fact that

  • molarity = (moles of solute)/(liters of solution)

1) For 175 mL of solution at 0.203 M, this means that:

  • 0.203 = (moles of solute)/0.175
  • moles of solute = 0.035523 mol

Considering the hydrochloric acid solution, if we have 0.035523 mol, then:

  • 6.00 = 0.035523/(liters of solution)
  • liters of solution = 0.035523/6.00 = 0.0059205 = <u>5.92 mL (to 3 sf)</u>

<u />

2) If there is 20.3 mL = 0.0203 L, then:

  • 8.20 = (moles of solute)/0.0203
  • moles of solute = 0.16646 mol

This means that the molarity of the diluted solution is:

  • 0.16646/(0.200) = <u>0.832 M (to 3 sf)</u>

<u />

3) If we need 1.50 L of 0.700 M solution, then:

  • 0.700 = (moles of solute)/1.50
  • moles of solute = 1.05 mol

Considering the 9.36 M acid solution, from which we need 1.05 mol of perchloric acid from,

  • 9.36 = 1.05/(liters of solution)
  • liters of solution = 1.05/9.36, which is 0.11217948717949 L, or <u>112 mL (to 3 sf)</u>
8 0
2 years ago
Heat required to raise 1 g of a substance 1°C
Oksana_A [137]

Answer:

Specific heat

Explanation:

7 0
3 years ago
Which of the following lists elements from lowest to highest atomic number
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4 years ago
how much energy would it take to heat a section of the copper tubing that weights about 660.0 gram, from 12.93 degree Celsius to
expeople1 [14]

Answer:

                      Q  =  2647 J

Explanation:

                    Specific heat capacity is the amount of energy required by one Kg of a substance to raise its temperature by 1 °C.

In thermodynamics the equation used is as follow,

                                                 Q  =  m Cp ΔT

Where;

           Q  =  Heat  =  ?

           m  =  mass  =  660 g

           Cp  =  Specific Heat Capacity  =  0.3850 J.g⁻¹.°C⁻¹

           ΔT  =  Change in Temperature  =  23.35 °C - 12.93 °C  =  10.42 °C

Putting values in eq. 1,

                            Q  =  660 g × 0.3850 J.g⁻¹.°C⁻¹ ×  10.42 °C

                            Q  =  2647 J

8 0
3 years ago
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